Question: What is the area of the region enclosed by the graphs of $f(x)=\sqrt{x+7}$ and $g(x)=0.5(x+7)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4$ (Choice B) B $ \dfrac{4}{3}$ (Choice C) C $\dfrac{52}{3}$ (Choice D) D $\dfrac{163}{12}-\dfrac{2}{3}\sqrt{343}$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${\llap{-}2}$ ${\llap{-}4}$ ${\llap{-}6}$ ${1}$ ${2}$ $f$ $g$ $y$ $x$ From the graph, it appears that $f(x)\ge g(x)$ between the points where the graphs intersect. From this we are looking to evaluate: $ \int_{a}^{b}\left( f(x)-g(x) \right)\,dx$ where $a$ and $b$ are the $x$ -coordinates of the points of intersection. Finding the $x$ -coordinates of the intersection points We can find the $x$ -coordinate of each point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{aligned} f(x)&=g(x) \\\\ \sqrt{x+7}&= 0.5(x+7)\\\\ x+7 &= 0.25(x+7)^2 \\\\ 0 &=0.25(x+7)^2 - (x+7) \\\\ 0 &= (x+7)(0.25(x+7)-1) \\ 0 &= (x+7)(0.25x+0.75) \end{aligned}$ We have two possible solutions at $x={-7}$ and $x={-3}$. Because this was a radical equation, let's check for extraneous solutions by plugging our possible solutions back into the original equation. $\begin{aligned} \sqrt{{-7}+7}&\stackrel{?}=0.5({-7}+7)\\\\ 0& \stackrel{\checkmark}= 0 \end{aligned}$ $\begin{aligned} \sqrt{{-3}+7}&\stackrel{?}=0.5({-3}+7)\\\\ 2& \stackrel{\checkmark}= 2 \end{aligned}$ The graphs intersect where $x=-7$ and $x=-3$. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $ \int_{-7}^{-3}\left(\sqrt{x+7}-(0.5(x+7))\right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{-7}^{-3} \left( \sqrt{x+7}- (0.5(x+7)) \right) \,dx \\\\ &= \dfrac{2}{3}(x+7)^{3/2}-\dfrac14x^2-\dfrac72x~\Bigg|_{-7}^{-3} \\\\ &= \left( \dfrac{16}{3}-\dfrac{9}{4}+\dfrac{21}{2} \right) - \left( 0 - \dfrac{49}{4}+\dfrac{49}{2}\right) \\\\ &= \dfrac{4}{3} \end{aligned}$ Answer The area is $ \dfrac{4}{3}$ square units.